Question: Simplify; express your answer in exponential form. Assume $r\neq 0, t\neq 0$. $\dfrac{{(r^{-1})^{5}}}{{(r^{-5}t)^{5}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${r^{-1}}$ to the exponent ${5}$ . Now ${-1 \times 5 = -5}$ , so ${(r^{-1})^{5} = r^{-5}}$ In the denominator, we can use the distributive property of exponents. ${(r^{-5}t)^{5} = (r^{-5})^{5}(t)^{5}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(r^{-1})^{5}}}{{(r^{-5}t)^{5}}} = \dfrac{{r^{-5}}}{{r^{-25}t^{5}}}$ Break up the equation by variable and simplify. $\dfrac{{r^{-5}}}{{r^{-25}t^{5}}} = \dfrac{{r^{-5}}}{{r^{-25}}} \cdot \dfrac{{1}}{{t^{5}}} = r^{{-5} - {(-25)}} \cdot t^{- {5}} = r^{20}t^{-5}$.